# Lecture 14: Random variables

• Reading: Cameron 3.1–3.3,
• Last semester's notes

• examples of Bayes' rule and the law of total probability
• definitions: random variable, expectation

## Law of total probability

Claim: (law of total probability) If $$A_1, \dots, A_n$$ partition the sample space $$S$$ (that is, if $$A_i \cap A_j = \emptyset$$ for $$i \neq j$$ and $$S = \cup A_i$$), then

$Pr(B) = \sum_{i} Pr(B|A_i)Pr(A_i)$

Proof sketch: Write $$B = \cup_{i} (B \cap A_i)$$. Apply third axiom to conclude $$Pr(B) = \sum_{i} Pr(B \cap A_i)$$. Apply definition of $$Pr(B | A_i)$$.

## Example using Bayes's rule and law of total probability

Suppose we are given a test for a condition. Let $$A$$ be the event that a patient has the condition, and let $$B$$ be the event that the test comes back positive.

The probability that a patient has the condition is $$Pr(A) = 1/10000$$. The test has a false positive rate of $$Pr(B | \bar{A}) = 1/100$$ (a false positive is when the test says "yes" despite the fact that the patient does not have the disease), and a false negative rate of $$Pr(\bar{B} | A) = 5/100$$.

Suppose a patient tests positive. What is the probability that they have the disease? In other words, what is $$Pr(A|B)$$?

Bayes's rule tells us $$Pr(A|B) = \frac{Pr(B|A)Pr(A)}{Pr(B)}$$. We can find $$Pr(B|A)$$ using the fact from last lecture: $$Pr(B|A) = 1 - Pr(\bar{B}|A) = 95/100$$. $$Pr(A)$$ is given. We can use the law of total probability to find $$Pr(B)$$; $$Pr(B) = Pr(B|A)Pr(A) + Pr(B|\bar{A})Pr(\bar{A})$$.

Plugging everything in, we have

\begin{aligned} Pr(A|B) &= \frac{Pr(B|A)Pr(A)}{Pr(B|A)Pr(A) + Pr(B|\bar{A})Pr(\bar{A})} \\ &= \frac{(95/100)(1/10000)}{(95/100)(1/10000) + (1/100)(9999/10000)} \\ &= \frac{95}{95+9999} \approx 1/100 \\ \end{aligned}

This is a surprising result: we take a test that fails $$\lt 5$$% of the time, and it says we have the disease, yet we have only about a 1% chance of having the disease.

However, note that our chances have grown from $$0.0001$$ to $$0.01$$, so we did learn quite a bit from the test.

## Random variables

Definition: A ($$\mathbb{R}$$-valued) random variable $$X$$ is a function $$X : S → \mathbb{R}$$.

Definition: The expected value of $$X$$, written $$E(X)$$ is given by $E(X) ::= \sum_{k \in S} X(k)Pr(\{k\})$

Definition: Given a random variable $$X$$ and a real number $$x$$, the poorly-named event $$(X = x)$$ is defined by $$(X = x) ::= \{k \in S \mid X(k) = x\}$$.

This definition is useful because it allows to ask "what is the probability that $$X = x$$?"

Claim: (alternate definition of $$E(X)$$) $E(X) = \sum_{x \in \mathbb{R}} x\cdot Pr(X=x)$

Proof sketch: this is just grouping together the terms in the original definition for the outcomes with the same $$X$$ value.

Note: You may be concerned about "$$\sum_{x \in \mathbb{R}}$$. In discrete examples, $$Pr(X = x) = 0$$ almost everywhere, so this sum reduces to a finite or at least countable sum. In non-discrete example, this summation can be replaced by an integral. Measure theory is a branch of mathematics that puts this distinction on firmer theoretical footing by replacing both the summation and the integral with the so-called "Lebesgue integral". In this course, we will simply use "$$\sum$$" with the understanding that it becomes an integral when the random variable is continuous.

Example: Suppose I roll a fair 6-sided die. On an even roll, I win \$10. On an odd roll, I lose however much money is shown. We can model the experiment (rolling a die) using the sample space $$S = \{1,2,3,4,5,6\}$$ and an equiprobable measure. The result of the experiment is given by the random variable $$X : S → \mathbb{R}$$ given by $$X(1) ::= -1$$, $$X(2) ::= 10$$, $$X(3) ::= -3$$, $$X(4) ::= 10$$, $$X(5) ::= -5$$, and $$X(6) ::= 10$$.

According to the definition,

\begin{aligned} E(X) &= (1/6)X(1) + (1/6)X(2) + (1/6)X(3) + (1/6)X(4) + (1/6)X(5) + (1/6)X(6) \\ &= (1/6)(-1) + (1/6)(10) + (1/6)(-3) + (1/6)(10) + (1/6)(-5) + (1/6)(10) \\ \end{aligned}

According to the alternate definition, $$E(X)$$ is given by

\begin{aligned} E(X) &= (-1)Pr(X = -1) + (-3)Pr(X = -3) + (-5)Pr(X = -5) + 10Pr(X = 10) \\ &= (-1)(1/6) + (-3)(1/6) + (-5)(1/6) + (10)(1/6 + 1/6 + 1/6) \end{aligned}

Definition: The probability mass function (PMF) of $$X$$ is the function $$PMF_X : \mathbb{R} → \mathbb{R}$$ given by $$PMF_X(x) = Pr(X = x)$$.